I'm new to Django and for my first project I'm building a portfolio. And I need a little kick-start with pagination help. I have an "index" view with a list of projects and a detail view of each project. In the detail view, I want a feature to be able to paginate between each individual object. I've gone through the Pagination documenation and applied what I learned with my index view but when I try to do the same thing with my detail view I get a TypeError saying my "object of type 'Project' has no len().
Here's a sample of my views.py code for reference:
def index( request ):
all_projects = Project.objects.all().order_by( '-pub_date' )
paginator = Paginator( all_projects, 12 )
try:
page = int( request.GET.get( 'page','1' ))
except ValueError:
page = 1
try:
projects = paginator.page( page )
except (EmptyPage,InvalidPage):
projects = paginator.page( paginator.num_pages )
return render_to_response( 'portfolio/index.html', { 'all_projects':all_projects, 'projects':projects, 'MEDIA_URL':MEDIA_URL })
def detail( request, project_id ):
project = get_object_or_404( Project, id=project_id )
return render_to_response( 'portfolio/detail.html', { 'project':project, 'MEDIA_URL':MEDIA_URL } )
Apologies if I sound n00b-ish because I am, and gratitude in advance for any help. Also, I read this previous post but it didn't seem to apply to me because my views aren't Class-based.
Django has built-in get_next_by_FOO()/get_previous_by_FOO() methods which will return the next/previous object depeding on a datetime-field.
You could access them in the template by somethow like:
{{ project.get_next_by_pub_date.title }}
I would say that this is the preferred method over using pagination for that, as you will get a nice url you can define in your model's get_absolute_url for every item!
To paginate you need an instance of a QuerySet, not an object! So you should replace your get_object_or_404 call by a filter/all. So it would be basically the same as the list view, but just pass the number 1 to the paginator, as you already do!
I actually do not have that setup through Django's pagination. I used this code:
prev_issue = Issue.objects.filter(title=title).filter(number__lt=issue.number).order_by('-number')[:1]
next_issue = Issue.objects.filter(title=title).filter(number__gt=issue.number).order_by('number')[:1]
In your case, I would do something like this, but you would have to filter by something, too:
prev_project = Project.objects.order_by('-pub_date')[:1]
next_project = Project.objects.order_by('pub_date')[:1]
Then put the two in the context.
I also recommend django-pagination. I notice you probably want the pagination to stay on the index, correct? http://code.google.com/p/django-pagination/
You just have to mess with the template codes and it works great.
The great thing about Django is, that there is code for nearly every common programming problem, that you can download and use in your own project. (and most of the django packages are quiet well written too)
for pagination check out http://pypi.python.org/pypi/django-pagination
it's really easy to install and setup, so you do not have to think about pagination and focus on coding your software!
hope this helps, Anton
In my views.py
def ProjectDetail(request,pk):
context = {}
template = 'views/projectdetail.html'
project = ''
prev = Project.objects.filter(pk__lt=pk).order_by('-pk')[:1]
next = Project.objects.filter(pk__gt=pk).order_by('pk')[:1]
try:
print(prev[0].pk)
print(next[0].pk)
except:
pass
project = Project.objects.filter(pk=pk)
context['categories'] = ProjectCategory.objects.all()
paginator = Paginator(project, 1) # Show 25 contacts per page
page = request.GET.get('page')
try:
data = paginator.page(page)
except PageNotAnInteger:
# If page is not an integer, deliver first page.
data = paginator.page(1)
except EmptyPage:
# If page is out of range (e.g. 9999), deliver last page of results.
data = paginator.page(paginator.num_pages)
if prev:
context['prev'] = prev[0].pk
if next:
context['next'] = next[0].pk
context['data'] = data
return render_to_response(template, context,context_instance=RequestContext(request))
in my template i have
<div class="row">
<a {% if next %} href="{% url 'task:project-detail' next %}" class="btn btn-primary pull-right" {% else %} class="btn btn-primary pull-right disabled" {% endif %}>Next</a>
<a {% if prev %} href="{% url 'task:project-detail' prev %}" class="btn btn-primary pull-left" {% else %} class="btn btn-primary pull-left disabled" {% endif %} >Previous</a>
</div>
Related
I am relatively new with Django, this must be a common problem.
I have created a view to show a form to input date (using widget that returns separate fields):
when date is inserted, I call a function userPage(request, my_date)
that filters, processes and renders a page (user.html) showing a list of items.
def datePage(request):
user=request.user
context = {}
context['form'] = UserDateForm()
if request.GET:
date_yr = request.GET['food_date_year']
date_mo = request.GET['food_date_month']
date_day = request.GET['food_date_day']
my_date_string = date_yr+'-'+date_mo+'-'+date_day
my_date = datetime.strptime(my_date_string, "%Y-%m-%d").date()
return userPage(request,my_date)
return render(request, "date.html", context)
def userPage(request, my_date):
user=request.user
# process context using user, my_date
context={...:..., 'my_date': my_date}
return render(request,'user.html',context)
In user.html I include a URL to add an item:
</div>
<form action="{% url 'My_ItemCreate' %}" method="POST">
{%csrf_token%}
<button type="submit" class="btn btn-success">
<span class="glyphicon glyphicon-plus"></span>
</button>
</form>
</div>
'My_ItemCreate' points to a django.views.generic CreateView that creates an item.:
path('MyItemCreate/',views.My_ItemCreate.as_view(),name='My_ItemCreate'),
class My_ItemCreate(CreateView):
model = MyItem
fields = ...
After creating the item in the CreateView, how do I go back to the user page
after I inserted the date? I have lost the date in the new URL.
If I use URL resolver to go to userPage, how do I pass a date in the format?
It would be nice that I am able to pass initial values in the CreateView, and
make some fields read-only, how do I modify/override CreateView ?
Many Thanks for your help!
I have found an answer to my problem: using request.session
to store a value and retrieving in other views, it works fine.
I am still curious to know if there are experts who
would provide a more elegant solution, and if someone
could be so kind to answer point 2) regarding CreateView read_only fields
Thanks
D
On my page all the blog posts will be showing up. So I wanted to implement a next/previous button to better my page.
def PostLists(request):
num = request.session.get('num',-5)
request.session['num'] = num
num = num + 5
exp = Post.objects.order_by('-date').all()[num:(num+5)]
context = {
'object_list': exp
}
if (request.user.is_authenticated):
return render(request, 'dashboard/postlist.html', context=context)
else:
return redirect('login')
I added a next button in my html code which will redirect me to my the same views.py function as shown above and the variable(num) shall increment by 5 thus showing me my next 5 posts. However this seems not to be working as I see the same 5 posts always.
Is there a better way to implement a next/previous button? If so could you please specify that? Thanks a lot!
I think you try to do too much yourself. Django has support for this, it even has a lot of support when rendering lists, and enforcing that the user is logged in.
We can use a class-based view for this: a ListView:
from django.contrib.auth.mixins import LoginRequiredMixin
from django.views.generic import ListView
class PostListView(LoginRequiredMixin, ListView):
model = Post
template_name = 'dashboard/postlist.html'
paginate_by = 5
queryset = Post.objects.order_by('-date')
In your dashboard/postlist.html template, then you can add logic to render the buttons. Like for example:
<!-- render the list -->
{% if is_paginated %}
{% if page_obj.has_previous %}
previous
{% endif %}
{% if page_obj.has_next %}
next
{% endif %}
{% endif %}
In the urls.py you can then use the PostListView.as_view() instead of the PostLists. So the ListView will here handle authentication check, slicing, pagination, etc.
I am trying out to build full-text searching by using sphinx search, postgresql & django based on this tutorial: http://pkarl.com/articles/guide-django-full-text-search-sphinx-and-django-sp/.
All setup done for sphinx & postgresql and it works but I got trouble when reach on Sample Django code part.
In django views & urlconf, I only changed the function of *search_results* into search and Story model with my own model. For URLConf, I only changed *search_results* into search just same like on views and nothing changed made on search template.
So when I try to search from my form in Django, I get exception:
TypeError at /search/
list() takes exactly 1 argument (0 given)
I also try to changed based on steyblind's comment by change the urlpattern & view definition like this:
(r'^search/(.*)?', search),
def search(request, query=''):
but still keep get TypeError exception.
Is there any mistake I am doing here? Thanks in advance.
Here's my snippets:
Urls.py
(r'^search/(.*)', search),
Views.py
def search(request, query):
try:
if(query == ''):
query = request.GET['query']
results = Flow.search.query(query)
context = { 'flows': list(results),'query': query, 'search_meta':results._sphinx }
except:
context = { 'flows': list() }
return render_to_response('search.html', context, context_instance=RequestContext(request))
search.html
{% extends "base.html" %}
{% block main %}
<div>
<form action="/search/" method="GET">
<input type="text" name="query"/>
<input type="submit">
</form>
{% if flows %}
<p>Your search for “<strong>{{ query }}</strong>” had <strong>{{ search_meta.total_found }}</strong> results.</p>
<p>search_meta object dump: {{ search_meta }}</p>
{% endif %}
<hr/>
{% for s in flows %}
<h3>{{ s.title }}</h3>
<p>(weight: {{ s.sphinx.weight }})</p>
<p>story.sphinx object dump: {{ s.sphinx }}</p>
{% empty %}
<p>YOU HAVEN'T SEARCHED YET.</p>
{% endfor %}
</div>
{% endblock %}
Correct me if I'm wrong, but Django-Sphinx seems to be an abandoned project. The last update to it was a year ago, with most updates being 3-5 years ago. Also, I cannot speak for Django then, but it can do now, out of the box, what you are trying to do with Sphinx.
What version of Django and Python are you using? The error you are getting is strange as list() can take no arguments. Try this in a python shell:
>> list()
[]
Regardless, I've made a few modifications to the code that could possibly help fix the issue. However, if there are no results, you are passing 'flows' as empty in this line:
context = { 'flows': list() }
If you look at the template, this really accomplishes nothing.
urls.py:
(r'^search/', search),
views.py:
def search(request):
query = request.GET.get('query')
if query:
results = Flow.search.query(query)
if results:
context = { 'flows': list(results),'query': query, 'search_meta':results._sphinx }
else:
context = { 'flows': list() }
return render_to_response('search.html', context, context_instance=RequestContext(request))
All that said, I'd highly suggest that since this project is so outdated that you use your own search. Or if you need more functionality, you could use a search app like Haystack which is updated frequently. Using the same urls.py as above, you could implement the below as an easy search that will return all results for a blank search, the actual filtered results for a query.
views.py:
def search(request):
query = request.GET.get('q')
results = Flow.objects.all()
if query:
results = results.query(query)
return render_to_response('search.html', {"flows": results,}, context_instance=RequestContext(request))
I'm using this code for my pagination, and I'd like the user's choice to be persistent throughout the site (this has been solved so far)...the only problem now is that the session variable now is permanent until the session is cleared by closing the browser. Also, how can I get the adjacent pages displayed...like in the digg-style Django paginator. I haven't been able to make sense of how to implement this into my code.
The code is as follows:
from django.core.paginator import Paginator, InvalidPage, EmptyPage
def paginate(request, object_list, paginate_by=10):
try:
if "per_page" in request.session:
per_page = request.session["per_page"]
else:
request.session["per_page"] = int(request.REQUEST['p'])
per_page = request.session["per_page"]
request.session.set_expiry(0)
except:
per_page = 10
paginator = Paginator(object_list, per_page)
try:
page = int(request.GET.get('page', '1'))
except ValueError:
page = 1
try:
items = paginator.page(page)
except (EmptyPage, InvalidPage):
items = paginator.page(paginator.num_pages)
return items
Then in my template I have this to render the pagination links:
<div class="pagination" align="center">
<span class="step-links">
{% if items.has_previous %}
previous
{% endif %}
<span class="current">
Page {{ items.number }} of {{ items.paginator.num_pages }}
</span>
{% if items.has_next %}
next
{% endif %}
</span>
</div>
You can achieve this by enabling sessions.
I recommend reading through the chapter Sessions, Users, and Registration on the djangobook website.
Edit: Now that you've enabled sessions, I think the problem is the hyperlinks in the template. Use an ampersand to separate multiple parameters in a url, for example
next
Edit 2: I'm not sure if I understand what the problem is with the session expiry. The line that sets the session to expire when the browser closes is request.session.set_expiry(0). See the django docs on Using Sessions in views if you want to change that.
To make a Digg style paginator, you need to write a function that takes the current page number and the total number of pages, and returns a list of page numbers. Then, in the template, loop through the page numbers and construct links to the pages.
A list of lists of page numbers would allow you to split the page numbers into groups, eg
[[1,2], [20,21,22,23,24], [30,31]]
I'm currently developing a Django application which will make use of the infamous "pagination" technique. I'm trying to figure out how the django.core.paginator module works.
I have an application with a Question model. I will be listing all of the questions using this paginator. There will be 20 questions per page.
def show_question(question_pk):
questions = Question.objects.all()
paginator = Paginator(questions, 20)
page = ... # Somehow figure out which page the question is on
return render_to_response('show_question.html', { 'page' : page })
In the view, where I list the different pages as "... 2, 3, 4, 5, 6, ..." I want to highlight the current page somehow, like many pages do.
There are really two things I want to know:
How do I make Django figure out which page the question is located at?
How would I write my template to properly "highlight" the currently visited page?
EDIT: Sorry, I forgot part of this question. I would also like any page except for the current one to be a link to /questions/{{ that_page.start_index }}. So basically every page link would link to the first question on that page.
Hmm... I see from your comment that you don't want to do the ol' GET parameter, which is what django.core.paginator was written for using. To do what you want, I can think of no better way than to precompute the page that each question is on. As an example, your view will end up being something like:
ITEMS_PER_PAGE = 20
def show_question(question_pk):
questions = Question.objects.all()
for index, question in enumerate(questions):
question.page = ((index - 1) / ITEMS_PER_PAGE) + 1
paginator = Paginator(questions, ITEMS_PER_PAGE)
page = paginator.page(questions.get(pk=question_pk).page)
return render_to_response('show_question.html', { 'page' : page })
To highlight the current page in the template, you'd do something like
{% for i in page.paginator.page_range %}
{% ifequal i page.number %}
<!-- Do something special for this page -->
{% else %}
<!-- All the other pages -->
{% endifequal %}
{% endfor %}
As for the items, you'll have two different object_lists to work with...
page.object_list
will be the objects in the current page and
page.paginator.object_list
will be all objects, regardless of page. Each of those items will have a "page" variable that will tell you which page they're on.
That all said, what you're doing sounds unconventional. You may want to rethink, but either way, good luck.
Django, at least from version 1.2, allows us to complete this task by using pure default pagination template tags.
{% for page in article_list.paginator.page_range %}
{% if page == article_list.number %}
{{ page }}
{% else %}
{{ page }}
{% endif %}
{% endfor %}
Where article_list is instance of
paginator = Paginator(article_list, 20)
try:
article_list = paginator.page(int(page))
except (EmptyPage, InvalidPage):
article_list = paginator.page(paginator.num_pages)
django-pagination should do what you want and comes wrapped in a pretty package you can just plug-in and use. It essentially moves the code from your views to the templates and a middleware.
EDIT: I just saw your edit.
You can get the current objects on a page using {% autopaginate object_list %}, which replaces object_list with the current objects for any given page. You can iterate through it and if you want the first, you should be able to treat it like a list and do object_list[0].
If you want to keep this within your views, you could do something like this:
def show_question(question_pk):
questions = Question.objects.all()
paginator = Paginator(questions, 20)
return render_to_response('show_question.html', { 'page' : paginator })
Within your template, you can access the current page you're on by doing:
# Gives you the starting index for that page.
# For example, 5 objects, and you're on the second page.
# start_index will be 3.
page.start_index
# You can access the current page number with:
# 1-based index
page.number
With that, you should be able to do everything you need.
There are a couple good examples here.